## Chemistry: The Central Science (13th Edition)

$[PH_3]$= 0.0432 mol/L $[BCl_3]$= 0.0432 mol/L
Kc = $\frac{products}{reactants}$ = $\frac{[PH_3][BCl_3]}{1}$ We denote the reactants as (1) because the reactant $PH_3$$BCl_3$$(s)$ is a solid. Solids and liquids are not included in equilibrium constants. Step 1: $Kc = [PH_3][BCl_3]$ Therefore: $1.87\times 10^{-3} = [PH_3][BCl_3]$ Step 2: In the reaction the mole ratio for the products $[PH_3][BCl_3]$ is one to one. Therefore: We can denote $[PH_3][BCl_3]$ as $[x][x]$. Step 3: $[x][x]$ = $x^{2}$ So: $1.87\times 10^{-3} = x^{2}$ Using algebraic laws for exponents, we see that by squaring both sides of the equation the answer is: $x = 0.0432$ $mol/L$ Rounded to three significant digits.